The sum of the first four terms of an A.P. is | vedclass

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(B) Let the first term be $a = 11$ and the common difference be $d$.The sum of the first four terms is given by: $a + (a + d) + (a + 2d) + (a + 3d) =

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$11$

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(B) Let the first term be $a = 11$ and the common difference be $d$.The sum of the first four terms is given by: $a + (a + d) + (a + 2d) + (a + 3d) = 56$.Substituting $a = 11$: $11 + (11 + d) + (11 + 2d) + (11 + 3d) = 56$.$44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.Let the number of terms be $n$. The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n$.These are $(11 + (n-4)2), (11 + (n-3)2), (11 + (n-2)2), (11 + (n-1)2)$.The sum of these terms is $112$:$44 + 2(n-4 + n-3 + n-2 + n-1) = 112$.$44 + 2(4n - 10) = 112$.$44 + 8n - 20 = 112$.$8n + 24 = 112$.$8n = 88 \Rightarrow n = 11$.

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,the number of terms is

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